What is the integral of 1/(x^2+64) ?

The integral of 1/(x^2+64) is 1/8 arctan(x/8)+C

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integral of 1/(x^2+64)

\int \frac{1}{x ^{2} + 64} d x

\frac{1}{8} arctan (\frac{x}{8}) + C

Solution steps

\int \frac{1}{x ^{2} + 64} d x

Apply Integral Substitution

= \int \frac{1}{8 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{8} arctan (u)

Substitute back

u = \frac{x}{8}

= \frac{1}{8} arctan (\frac{x}{8})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{x}{8}) + C

y = 2 x^{3} - x^{2} + 2

\int \frac{e ^{x^{n}}}{x ^{1 - n}} d x

h \to 0 lim (\frac{5 e ^{x} - 5 e ^{x + h}}{3 h})

\frac{e ^{- s}}{( s ^{2} + 8 )}

\frac{d}{d x} (\frac{3 x - 1}{2 x + 1})

What is the integral of 1/(x^2+64) ?
The integral of 1/(x^2+64) is 1/8 arctan(x/8)+C